Calculus I Quiz 4

Calculus I Quiz 4 解答修正版 Dec. 29, 2008

Name: Student ID number:

1. (20 pts; 2 pts for each problem) Find the general antiderivative of the given function (f (x) = F⁰(x)).

• For x6= 0, f(x) = 1

2x, F (x) = ¹₂ln|x| +C

• f (x) = x², F (x) = ¹₃x³ +C

• f (x) = (x + 1)², F (x) = ¹₃(x + 1)³ +C

• f (x) = e^4x, F (x) = ¹₄e^4x +C

• f (x) = 3x^1/3, F (x) = ⁹₄x^4/3 +C

• f (x) = e^x+ x, F (x) = e^x+¹₂x² +C

• f (x) = 1, F (x) = x +C

• f (x) = cos x, F (x) = sin x +C

• f (x) = 1

1 + x², F (x) = tan⁻¹x +C

• For x6= 0, f(x) = x²+ 1

x , F (x) = ¹₂x²+ ln|x| +C

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Name: Student ID number:

2. (10 pts) The sequence {xn} is recursively defined.

x_n+1 = 2xn

1 + x_n

• (2 pts) Find all equilibria (fixed points) of {xn}.

x = 2x

1 + x ⇒ x = 0, 1

• (8 pts)Determine the stability of the equilibria (fixed points).

f (x) = 2x

1 + x = 2− 2

1 + x ⇒f⁰(x) = 2 (1 + x)²

f⁰(0) = 2,⇒ |f⁰(0)| > 1 ⇒ x^∗ = 0 is unstable f⁰(1) = 1/2,⇒ |f⁰(1)| < 1 ⇒ x^∗ = 1 is stable

Note: f⁰(1) = 1/2 > 0, the equilibrium is approached without oscillations.

3. (10 pts) Find the derivative of

∫ x³ x²

tan t dt

Let F (x) =∫_x

0 tan t dt and G(x) =∫_x³

x² tan t dt = F (x³)− F (x²).

Since tan t is a continuous function, by FTC, F (x) is differentiable and F⁰(x) = tan x.

By Chain rule, G⁰(x) = F⁰(x³)· 3x²− F⁰(x²)· 2x = 3x²tan x³− 2x tan x².

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Name: Student ID number:

4. (10 pts)

∫ ₃

0

x dx

• Use a graph to interpret the definite integral in terms of the area.

• Find the value of the integral.

Area= ¹₂ · 3 · 3 = ⁹₂.

5. (10 pts) Find the area of the region bounded by the lines and curves y = x²+ 1, y = 4x− 2

{ y = x²+ 1,

y = 4x− 2 ⇒intersection points: (1, 2), (3, 10) Area=∫3

1[(4x− 2) − (x²+ 1)] dx = (−¹₃x³+ 2x² − 3x)|³₁ = ⁴₃. or Area=∫10

2 [√

y− 1 − (¹₄y + ¹₂)] dx =· · · = ⁴₃

6. (10 pts) Evaluate the indefinite integral

∫ 1

xe^{1+ln x}dx

Hint: Let t = 1 + ln x.

Ans: e^{1+ln x}+ C

另解: e^{1+ln x} = e¹· e^{ln x} = e¹x ...

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Name: Student ID number:

7. (10 pts) Evaluate the definite integral

∫ e² e

1 x(ln x)² dx

Let t = ln x, dt = ¹_xdx

∫ e² e

1

x(ln x)² dx =

∫ 2 1

1

t²dt = −1 t|²1 = 1

2

8. (10 pts) Evaluate the indefinite integral

∫

x³ln (x²+ 1) dx

Let t = x²+ 1, dt = 2x dx. ∫

x³ln (x²+ 1) dx =∫ ₁

2(t− 1) ln t dt.

Let u = ln t, dv = ¹₂(t− 1). Then du = ¹_tdt, v = ¹₄t²− ¹₂t

∫

x³ln (x²+ 1) dx

=

∫ 1

2(t− 1) ln t dt

=(1 4t²−1

2t) ln t−

∫ 1 4t−1

2dt

=(1 4t²−1

2t) ln t− 1 8t²+1

2t+C

=(1

4(x²+ 1)²− 1

2(x² + 1)) ln (x²+ 1)− 1

8(x² + 1)²+1

2(x²+ 1)+C

9. (10 pts) Evaluate the definite integral

∫ _π/6

0

e^xcos x dx

Hint: Sec 7.2, example 7.

Ans:

e^π/6(1 +√ 3 4 )−1

2

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